The expression is equivalent to for all real values of . What is the value of ?

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SAT Equivalent Expressions Question 186 | Free SAT Prep | Aniko

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Question 186·Hard·Equivalent Expressions

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The expression $(x^{2}+ax+b)(x+3)+(4x+c)$ is equivalent to $x^{3}+10x^{2}+19x+12$ for all real values of $x$ . What is the value of $b+c$ ?

For problems where two polynomial expressions are stated to be equivalent for all real $x$ , immediately expand the more complicated side, collect like terms, and write it in standard form. Then match coefficients of each power of $x$ to form simple linear equations in the unknowns and solve them in a logical order, starting from the easiest equation (often the highest or lowest degree). Finally, compute whatever combination of variables the question asks for, rather than solving for any extra quantities you don’t need.

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Think about what “equivalent for all x” implies

If two polynomial expressions are equal for all real values of $x$ , what does that tell you about the coefficients of $x^{3}$ , $x^{2}$ , $x$ , and the constant terms on each side?

Expand before comparing

First, expand $(x^{2}+ax+b)(x+3)$ carefully, then add $(4x+c)$ and combine like terms so the left side is written as a standard cubic $x^{3} + (\text{something})x^{2} + (\text{something})x + (\text{constant})$ .

Match coefficients to make equations

Once you have the left side in standard form, match its $x^{2}$ coefficient to $10$ , its $x$ coefficient to $19$ , and its constant term to $12$ to create equations involving $a$ , $b$ , and $c$ .

Solve in a logical order

Use the simplest equation (from the $x^{2}$ coefficient) to find $a$ first, then substitute to find $b$ , then use the constant-term equation to find $c$ , and finally compute $b + c$ .

Desmos Guide

Graph both sides as functions

In Desmos, enter the left side as

f(x) = (x^2 + a*x + b)*(x + 3) + (4*x + c)

and the right side as

g(x) = x^3 + 10*x^2 + 19*x + 12.

When you type a, b, and c, Desmos will offer to create sliders for them—accept this.

Adjust sliders to match the graphs

Move the sliders for $a$ , $b$ , and $c$ until the graph of $f(x)$ lies exactly on top of the graph of $g(x)$ for all visible $x$ -values. When the two graphs coincide everywhere, the slider values of $b$ and $c$ are the ones that make the expressions equivalent.

Compute b + c from the slider values

Once the graphs match perfectly, read the values of $b$ and $c$ from the sliders and add them to find $b + c$ . You can type b + c into Desmos to have it calculate the sum; that value should match one of the answer choices.

Step-by-step Explanation

Use the idea of matching coefficients

The expression

(x^{2}+ax+b)(x+3)+(4x+c)

is equal to $x^{3}+10x^{2}+19x+12$ for all real $x$ . That means when we expand and simplify the left side, the coefficients of $x^{3}$ , $x^{2}$ , $x$ , and the constant term must match the corresponding coefficients on the right side.

Expand the product

First expand $(x^{2}+ax+b)(x+3)$ :

\begin{aligned} (x^{2}+ax+b)(x+3) &= x^{2}(x+3) + ax(x+3) + b(x+3) \\ &= (x^{3} + 3x^{2}) + (ax^{2} + 3ax) + (bx + 3b) \end{aligned}

Now combine like terms:

(x^{2}+ax+b)(x+3) = x^{3} + (3+a)x^{2} + (3a + b)x + 3b.

Add the remaining terms and collect like terms

Now add the $(4x+c)$ part:

(x^{2}+ax+b)(x+3) + (4x+c) = x^{3} + (3+a)x^{2} + (3a + b)x + 3b + 4x + c.

Combine the $x$ terms and the constants:

= x^{3} + (3+a)x^{2} + (3a + b + 4)x + (3b + c).

So the left side, in standard form, is

x^{3} + (3+a)x^{2} + (3a + b + 4)x + (3b + c).

Set up equations by matching coefficients

Match the coefficients of each power of $x$ with those in $x^{3}+10x^{2}+19x+12$ :

Coefficient of $x^{3}$ : $1$ on both sides (already matches).
Coefficient of $x^{2}$ : $3 + a = 10$ .
Coefficient of $x$ : $3a + b + 4 = 19$ .
Constant term: $3b + c = 12$ .

Now solve these equations step by step.

Solve for a, then b, then c, and find b + c

From $3 + a = 10$ :

a = 7.

Substitute $a = 7$ into $3a + b + 4 = 19$ :

\begin{aligned} 3(7) + b + 4 &= 19 \\ 21 + b + 4 &= 19 \\ 25 + b &= 19 \\ b &= -6 \end{aligned}

Now use $b = -6$ in $3b + c = 12$ :

\begin{aligned} 3(-6) + c &= 12 \\ -18 + c &= 12 \\ c &= 30 \end{aligned}

Finally, compute $b + c$ :

b + c = -6 + 30 = 24

so the correct answer is $24$ (choice D).

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Step-by-step Explanation

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